$\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\const}{\mathrm{const}}$
Analyzing Example 11.6 and Example 11.7
If you have the Strauss textbook, it is discussed in quite a bit of detail in chapter 4.3. Except Strauss in his usual manner to tell truth, only truth, but never all the truth nixes the case of 2 negative eigenvalues as he assumes that either $\alpha>0$ or $\beta>0$.
The easiest way to deal with it would be to note that $\alpha+\beta+\alpha\beta l=0$ has two branches and divides plane into 3 regions and due to continuity of eigenvalue in each of them the number of negative eigenvalues is the same.
Consider $\beta=\alpha$, it transects all three regions. Shift coordinate $x$ to the center of interval, which becomes $[-L,L]$, $L=l/2$. Now problem becomes \begin{align} &X''+\lambda X=0,\label{equ-I.1}\\ &X'(-L)=\alpha X(-L),\label{equ-I.2}\\ &X'(L)=-\alpha X(L)\label{equ-I.3} \end{align} and therefore if $X$ is an eigenfunction, then $Y(x):=X(-x)$ is eigenfunction with the same eigenvalue.
Therefore we can consider separately eigenfunctions which are even functions, and which are odd functions--and those are described respectively by \begin{align} &X''+\lambda X=0,\label{equ-I.4}\\ &X'(0)=0,\label{equ-I.5}\\ &X'(L)=-\alpha X(L)\label{equ-I.6} \end{align} and \begin{align} &X''+\lambda X=0,\label{equ-I.7}\\ &X(0)=0,\label{equ-I.8}\\ &X'(L)=-\alpha X(L).\label{equ-I.9} \end{align} Since we are looking at $\lambda=-\gamma^2$ ($\gamma>0$, we look at $X=\cosh (x \gamma)$ and $X=\sinh (X\gamma)$ respectively (see conditions (\ref{equ-I.5}), (\ref{equ-I.8})) and then conditions (\ref{equ-I.6}), (\ref{equ-I.9}) tell us that \begin{align} &\alpha L = -(L\gamma)\tanh (L\gamma),\label{equ-I.10}\\ &\alpha L= - (L\gamma) \coth (L\gamma)\label{equ-I.11} \end{align} respectively.
Both functions $z\tanh(z)$ and $z\coth(z)$ are monotone increasing for $z>0$ with minima at $z=0$ equal $0$ and $1$ respectively. Thus equation (\ref{equ-I.10}) has a single solution $\gamma$ iff $\alpha<0$ and (\ref{equ-I.11}) has a single solution $\gamma$ iff $\alpha L < -1$.
Therefore as $\alpha l<0$ there is one negative eigenvalue with an even eigenfunction and as $2\alpha l+(\alpha l)^2<0$ comes another negative eigenvalue with an odd eigenfunction.
Sure, one can apply a variational argument (see Appendix H): but analysis above has its own merit (mainly learning).